Divide using synthetic division $$ ( 9x^{3}+3x^{2}+10x-1 ) \div ( 3x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&9&3&10&-1\\& & 9& 12& \color{black}{22} \\ \hline &\color{blue}{9}&\color{blue}{12}&\color{blue}{22}&\color{orangered}{21} \end{array} $$The solution is:
$$ \frac{ 9x^{3}+3x^{2}+10x-1 }{ x-1 } = \color{blue}{9x^{2}+12x+22} ~+~ \frac{ \color{red}{ 21 } }{ x-1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{1}&9&3&10&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 9 }&3&10&-1\\& & & & \\ \hline &\color{orangered}{9}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 9 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&9&3&10&-1\\& & \color{blue}{9} & & \\ \hline &\color{blue}{9}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 9 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}1&9&\color{orangered}{ 3 }&10&-1\\& & \color{orangered}{9} & & \\ \hline &9&\color{orangered}{12}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 12 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&9&3&10&-1\\& & 9& \color{blue}{12} & \\ \hline &9&\color{blue}{12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 12 } = \color{orangered}{ 22 } $
$$ \begin{array}{c|rrrr}1&9&3&\color{orangered}{ 10 }&-1\\& & 9& \color{orangered}{12} & \\ \hline &9&12&\color{orangered}{22}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 22 } = \color{blue}{ 22 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&9&3&10&-1\\& & 9& 12& \color{blue}{22} \\ \hline &9&12&\color{blue}{22}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 22 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrr}1&9&3&10&\color{orangered}{ -1 }\\& & 9& 12& \color{orangered}{22} \\ \hline &\color{blue}{9}&\color{blue}{12}&\color{blue}{22}&\color{orangered}{21} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 9x^{2}+12x+22 } $ with a remainder of $ \color{red}{ 21 } $.